Final answer:
The frequency f of small-angle oscillations for a uniform rod swinging as a pendulum with pivot at a distance l/4 from one end is given by f = 1/(2π√(4∙l/3g)), using the formula for the period of a physical pendulum.
Step-by-step explanation:
To find the expression for the frequency f of small-angle oscillations for a uniform rod swinging as a pendulum, we can consider it as a physical pendulum. The period T of a physical pendulum can be found using the formula:
T = 2π√(I/mgL),
where I is the moment of inertia of the rod about the pivot, m is the mass of the rod, g is the acceleration due to gravity, and L is the distance from the pivot to the rod's center of mass. For a uniform rod of length l, the center of mass is at its midpoint, so L = l/4 in this case. The moment of inertia I for a rod about an end is (m∙l^2)/3.
Inserting these values, we get:
T = 2π√((m∙l^2)/3) / (m∙g∙(l/4)),
Which simplifies to:
T = 2π√(4∙l/3g).
The frequency f is the inverse of the period T, thus:
f = 1/T = 1/(2π√(4∙l/3g)).
This formula provides the frequency of small-angle oscillations for the given system.